Let $S = \{(a_1, \ldots , a_c) \ : \ \sum_i i a_i = n\}$, and for $a \in S$, let $f(a) = 1/\prod_i a_i !$.  Then the exact value you want is $n! \sum_{a \in S} f(a)$ [the term $n! f(a)$ counts the number of such decompositions with $a_i$ sets of size $i$].

We can bound this by finding some $f(a) \leq \Delta$, which would give

$$\Delta  \leq N / n! \leq |S| \Delta = p_c (n) \Delta$$

where $N$ is the number you want, and $p_c (n)$ is the number of partitions of $n$ into parts of size at most $c$.  (Then use Stirling's formula and some useful upper bound on like perhaps $p_c (n) \leq p(n) \sim \frac{1}{4n \sqrt{3}} \exp[\pi \sqrt{2n/3}].$)

Is that good enough?  If not, you could get more mileage out of these bounds.